Equation problem of straight lines and circles in high school mathematics

Let the coordinates of the center of the circle be (x,y).

Then find the distance from the points (-2,0) and (6,0) to the center of the circle.

Use the distance from these two points to the center of the circle to think of an equicolumn equation.

x=2 for 2-x) +0-y) = (6-x) +0-y) solution

Then substitute x=2 into 3x+2y=0

The solution is y=-3

The coordinates of the center of the circle are (2,-3).

Substituting the coordinates of the center of the circle and the coordinates of one point on the circle into the square + (y1-y2) under the root number (x1-x2) to get a radius of 5

Finally, it is substituted into the standard equation of the circle.

x-a）²+y-b）²=r²

(x-2) +y+3) =25

A system of equations can be solved.

First of all, the center of the circle is (x,y).

Then the center of the circle to the point (-2,0) is the radius, and the same to (6,0) is the radius.

Therefore, (x+2)*(x+2)+y*y=(x-6)*(x-6)+y*y, solve x=2, and then solve y=-3 according to 3x+2y=0, and then substitute (x+2)*(x+2)+y*y, and the radius is 5, and the equation is (x-2)*(x-2)+(y+3)*(y+3)=25

Because the intersection point with the x-axis is (-2,0)(6,0), where the perpendicular equation is x=2, and the center of the circle must be on two straight lines, so the simultaneous 3x+2y=0 and x=2

The coordinates of the center of the circle are (2,-3).

So the radius r=5 (using the point-to-point distance formula).

So the circle is (x-2) square + (y+3) square = 25

From the title, the center of the circle is also on x=2, and the coordinates of the center of the circle are (2,-3), and the radius of the same circle can be obtained by a point of coordinates, and the circle coordinates are (x-2) squared plus (y+3) squared=25

This is actually not difficult, after determining the coordinates and radius of the center of the circle, draw it in the coordinate system, and then draw the point b, the corresponding position relationship is clear at a glance, the rightmost side of the circle is just (, the top edge of the circle is just (so the tangent position is easy to find.

Let the tangent slope be k first, then the tangent equation is.

y-2 k(x-2), i.e. kx-y-2k+2 0

The center of the circle is (1,1) and the radius r is 1

So the distance from the center of the circle to the tangent.

d k-1-2k+2 k +1) r 1, i.e. k-1 k +1) 1, squared on both sides.

k-1）²＝k²+1

i.e. -2k 0The solution is k 0

So the tangent equation is y-2 0

And because the point b is a point outside the circle, there must be two tangents, so the other tangent must be slopeless, so the other tangent is x 2

If one of the two tangents is x=2 and the other has a slope of k, the distance from the center of the circle to the tangent is 1, which can be solved by using the point-line distance formula.

Let the equation of the straight line tangent to the circle through b(2,2) be y+2=k(x+2), that is, kx-y+2k-2=0, and the distance from the center of the circle (1,1) to the tangent is equal to the radius of the circle 1.

k-1-2k-2|k +1) = 1, and k = -4 3, then the equation is .

4 3*x-y+2*(-4 3)-2=0, i.e., 4x+3y+14=0,

(x-1)^2 +(y-1)^2 =1

The center of the circle c=(1,1), the radius r=1

Crossing the point b(2,2), the linear equation l:

y-2=m(x-2)

mx-y +(2-2m)=0

The distance from the center of the circle c=(1,1) to the linear equation l:mx-y +(2-2m)=0 =r

m-1 +(2-2m)|/m^2+1) =1|-m+1| =m^2+1)

m+1)^2 =(m^2+1)

m=0 tangent equation : y=2

Another tangent equation : x=2

Solution: Because the circle (x-1)2+(y-1)2=1

The center of the circle is (1,1), the radius is 1, and the point b(2,2), so the two tangents of the circle passing through the point b are x=2 and y=2 respectively.

1.Solution: X+2Y-3=0 and X+Y+X-6Y+M=0, remove X, 5Y-20Y+M+12=0, =400-20(M+12)>0Get.

m<8, let p(x1,y1), q(x2,y2), then y1+y2=4, y1y2=(12+m) 5, so x1+x2=6-2(y1+y2)=-2, x1x2=9-6(y1+y2)+4y1y2=(4m-27) 5, and pq is the diameter of the circle through the origin o, so the vector op oq = 0, that is, x1x2 + y1y2 = (4m-27) 5 + (12 + m) 5 = 0

The solution is m=3

2.Let the equation for the circle be (x-a) +y-b) =r and r=|a|. a-3b=0。。。

The distance from the center of the circle to the line y=x is |a-b|2, again known by the vertical diameter theorem, a -7 = (|a-b|2), i.e., a +2ab-b -14 = 0

Composed. A = 3 7 and b = 7, so the equation for the circle is (x-3 7) +y- 7) =63 or (x+3 7) +y+ 7) =63.

The simplest way to do this is to understand the circular equations.

The equation of a simultaneous line and a circle, x 2 + y 2-8y + m + 3 = 0 (the key is to understand this equation, x, y is the result of the synthesis, and must pass through the intersection of the line and the circle, which is again an equation of a circle, and therefore an equation of a circle through p and q).

The circle passes through the origin, and (0,0) is substituted to obtain m=-3, which will be more troublesome in other ways.

Didn't notice that pq is the diameter, I'm looking.

That is true. This equation has a bent and hypothetical vertical variable m. Therefore, this fixed point must have nothing to do with the burying of variables.

Rephrase this equation to this form: (x+y-3)m+2x-3y-16=0. After that, let x+y-3=0 and 2x-3y-16=0!

This way the equation becomes 0 times m plus zero equals zero, regardless of the value of m! The system of equations x=5,y=-2!Therefore it must pass the fixed point (5, 2), hehe!

In this problem, the fixed-point problem is to say that the point found has nothing to do with the value of m, and the equation must be true if it is substituted.

So the first step is to dissolve the left side of the equation.

Concentrate m: mx+2x+my-3y-3m-16=0 and extract m:

m（x+y-3）+2x-3y-16=0

To make the point that is found has no relationship with m as the fixed liquid wax point, x+y-3=02x-3y-16=0

Simultaneous inequality is solved to x=5

y=-2 is (5,-2).

If you still have questions, keep an eye on the trouble and continue to ask!

Solution: After the original line is deformed, it is:

m(x+y-3)+2x-3y-16=0

Just let x+y-3=0, 2x-3y-16=0

Then the equation of the straight line is true, that is, it has nothing to do with m, and it is over the vertex at this time.

Xie Fang Yu Shicheng group information elimination x+y-3=0,2x-3y-16=0., yield: x=5, y=-2

i.e. across the vertex (5,-2).

Equal difference series, a+b=2c

c=2, a+b=4, i.e. |ca|+|cb|=4, so the trajectory of point c is an ellipse with a and b as focal lengths.

2c=2c=1b*b+c*c=a*a

b*b+1=a*a

a+b=4 is obtained from the above equation a*a=4, b*b=1

The equation is x*x 4+y*y 3=1

1.(1) As shown in the figure (the coordinate system is omitted), the center of the circle n(-1,-1) is the midpoint of the chord AB, and in RT AMN, AM|2=|an|2+|mn|2,(m+1)2=-2(n+2).(

Therefore, the trajectory equation for the center m of the moving circle is (x+1)2=-2(y+2)

2) From (*), we know that (m+1)2=-2(n+2) 0, so there is n -2

And the radius of the circle m r= when r=

, n=-2, m=-1, and the equation for the circle is (x+1)2+(y+2)2=5

2..It is known that the circle o: x 2 + y 2 = 1 and the parabola y = x 2-2 are three different points a, b, cIf both the line ab and ac are tangent to the circle o. Verification: The straight line BC is also tangent to the circle O.

Using the parabolic equation, we can assume a(x1,x1 2-2).

b(x2x2^2-2)

c(x3,x3 2-2) then we can write ab

ACBC equations are then used (

00) to ab

AC is 1 to get two x1s

x2x3 and then we find (0

0) to bc, as long as the previously obtained formula is descended into this distance, the distance will also be 1, then the result is proved.

It is known that in a triangle, the edges of the angles a, b, and c are a, b, and c, respectively, and a>c>b are in a series of equal differences, |ab|2. Find the trajectory equation of the vertex.

Solution: a, c, b are equal difference series, c is set to the middle term, so c=(a+b) 2=|ab|＝2，∴a+b=4.

The coordinate system is established with the line where AB is located as the X-axis, and the midpoint of the line segment AB as the coordinate origin, in which the coordinates of a, b, and c are: a(-1,0); b(1，0)，c(x，y)；Then the trajectory of vertex c is an elliptical with a and b as the focus; Let the major semi-axis of the ellipse.

is m, the short half axis is n, f is the half focal length, (this is set to avoid confusion with a, b, c above), then 2m = 4, m = 2, f = 1, n = m -f = 4-1 = 3, so the trajectory equation for vertex c is.

x²/4+y²/3=1.

1.Let the equations of two straight lines be ax+by+c=0, ax by c=0, and their intersection points are (x0, y0), then.

ax0+by0+c=0

ax0＋by0＋c=0

So ax0+by0+c+n(ax0 by0 c)=0

Obviously, the straight line ax+by+c+n(ax by c)=0 is the passing point (x0, y0), and every value of n it represents a straight line, so it weighs the linear system equation of the intersection of two straight lines, when n=0 represents the first straight line, but it does not contain the second straight line, pay attention to the test when using.

2.The distance between two points is derived using the formula.

Let the intersection point of the line and the circle be a(x1,y1),b(x2,y2), when the slope of the line does not exist, find the coordinates of the intersection point, and directly find the chord length;

When the slope of the straight line exists, then.

Slope k=(y1-y2) (x1-x2).

ab|=√[(x1-x2)²+y1-y2)²|

[(x1-x2)²+y1-y2)²]

(x1-x2)² x1-x2)²+y1-y2)²]/√(x1-x2)²

(x1-x2)² x1-x2)²+y1-y2)²]/(x1-x2)²

(x1-x2)² 1+(y1-y2)²/(x1-x2)²]

(x1-x2)² 1+k²)

(1+k²)*x1+x2)² 4x1x2]

The center of the circle is O, then the area of the quadrilateral PAOB is equal to 2 times the area of the triangle Pao, and the triangle Pao is a right-angled triangle, and its area is (1 2) Oa Pa, considering the right-angled side of the right-angled triangle Oa=R=1, then the minimum value of the area of the triangle only needs to make the hypotenuse po small, the position of the point P is the projective point of the point O on the straight line, the distance from the point O to the straight line is d=10 13, then the minimum value of the Pa is (87) (13), Then the minimum value of the triangle PAO area is (87) (2 13), and thus the minimum value of the quadrilateral PAOB area is (87) (13).

x 2 + y 2 = 1 to get the center of the circle (

2x+3y+10=0 to get d=10 root number13 d>r The line and the circle are separated, and when the PO line segment is the smallest, there is a minimum value, that is, the line segment PO=D=10 The root number 13 tangent point a b is perpendicular to the radius.

s=s(oap)+s(obp)=1 2*87 13+1 2*87 13=87 13=87 13 in root